数列推导设等差数列An的前n项和为Sn,则S4,S8-S4,S12-S8,S16-S12成等差数列,设等差数列An的前n项和为Sn,则S4,S8-S4,S12-S8,S16-S12成等差数列,类比以上结论有:设等比数列Bn的前n项积为Tn,则T4,T8/T4,T12/T8,T16/T

Question

数列推导设等差数列An的前n项和为Sn,则S4,S8-S4,S12-S8,S16-S12成等差数列,
设等差数列An的前n项和为Sn,则S4,S8-S4,S12-S8,S16-S12成等差数列,
类比以上结论有:
设等比数列Bn的前n项积为Tn,则T4,T8/T4,T12/T8,T16/T12成等比数列,谁能证明一下啊.
证4,T8/T4,T12/T8,T16/T12成等比数列啦、是不是只要证出公比相等就可以

Answer 1

a5=a1+4d,a6=a2=4d...a(n+4)=a(n)+4d
(S8-S4)-S4=4d [S(4n+4)-S(4n)]=4d
同理可得S4,S8-S4,S12-S8,S16-S12成等差数列
t5=t1*q^4,t6=t2*q^4...t(n+4)=t(n)*q^4
(T8/T4)/T4=q^4 [T(4n+4)/T(4n)]=q^4
同理可得T4,T8/T4,T12/T8,T16/T12成等比数列